Random maps with unconstrained genus Thomas Budzinski Joint work with Nicolas Curien and Bram Petri ENS Paris July 16th 2019 Saint-Flour Probability Summer School Thomas Budzinski Random maps with unconstrained genus
General maps A rooted map is a gluing of polygons which yields a connected, orientable surface. It is equipped with a distinguished oriented edge. We do not assume that this surface is the sphere. There are finitely many maps with n edges. Question: let M n be a uniform map with n edges. What does it look like? Thomas Budzinski Random maps with unconstrained genus
Encoding maps via permutations Add labels on the half-edges (1 is the label of the root edge). 5 2 τ = ( 17 )( 23 )( 68 )( 59 )( 4 10 ) 6 σ = ( 1 8 9 4 10 )( 256 )( 37 ) 9 4 8 3 1 7 10 A map m with n edges can be encoded by a pair ( τ, σ ) of permutations of { 1 , 2 , . . . , 2 n } . τ is an involution with no fixed point that matches the two halves of each edges. σ describes how to turn around a vertex, and may be any permutation. Thomas Budzinski Random maps with unconstrained genus
Connectedness Small issue: a pair ( τ, σ ) does not always give a map (a map must be connected). However, the proportion of pairs ( τ, σ ) that correspond to a map goes to 1 as n → + ∞ . Consequence: the number of maps with n edges is asymptotically ( 2 n − 1 )!! × ( 2 n )! = 2 n × ( 2 n − 1 )!! . ( 2 n − 1 )! Other consequence: a uniform map M n is the map given by a uniform matching τ n and an independent uniform permutation σ n on { 1 , 2 , . . . , 2 n } , conditioned on an event of probability 1 − o ( 1 ) . Thomas Budzinski Random maps with unconstrained genus
Basic properties of M n The vertices of M n correspond to the disjoint cycles of the uniform permutation σ n . The number of cycles has the distribution of 2 n � 1 � � Ber = ( 1 + o ( 1 )) log n . i i = 1 By duality we also have # F ( M n ) = ( 1 + o ( 1 )) log n . By the Euler formula ( f + v − e = 2 − 2 g ), the genus of M n is n 2 − log n + o ( log n ) . This is close to the maximal possible value n 2 . Thomas Budzinski Random maps with unconstrained genus
Vertex degrees of M n The degrees of the vertices are the lengths of the cycles of σ n . The length of the cycle containing 1 in σ n has uniform distribution in { 1 , 2 , . . . , 2 n } . Conditionally on this cycle C n 1 , the rest of σ n is a uniform permutation of the complementary of C n 1 . Poisson–Dirichlet process: let ( X i ) i ≥ 1 decreasing re-ordering of the variables U 1 , ( 1 − U 1 ) U 2 , ( 1 − U 1 )( 1 − U 2 ) U 3 , . . . , where the U i are i.i.d. uniform variables on [ 0 , 1 ] . Let v n 1 , v n 2 , . . . be the vertices of M n , ranked by decreasing degrees. Then � 1 � ( d ) 2 n deg ( v n i ) − n → + ∞ ( X i ) i ≥ 1 . − − − → i ≥ 1 Thomas Budzinski Random maps with unconstrained genus
Configuration model Let G ( m ) be the graph associated to a map m . Conditionally on σ n , the graph G ( M n ) is given by the uniform matching τ n : it is a configuration model . Let [ i , j ] n be the number of edges joining v n i and v n j , with the convention that [ i , i ] n is twice the number of loops on v n i . Then � 1 � ( d ) 2 n [ i , j ] n − n → + ∞ ( X i X j ) i , j ≥ 1 . − − − → i , j ≥ 1 Thomas Budzinski Random maps with unconstrained genus
More general models We want to deal with more general models like triangulations, maps with one face... We start with a family P n of polygons with 2 n edges in total, and glue the edges two by two in a uniform way to obtain a map M P n . How many vertices in M P n ? Law of the largest degrees? Thomas Budzinski Random maps with unconstrained genus
Universality General idea: the graph G ( M P n ) looks a lot like G ( M n ) . Conjecture Assume that P n has o ( √ n ) one-gons and o ( n ) two-gons. Then the total variation distance between the laws of G ( M P n ) and G ( M n ) goes to 0 as n → + ∞ . The assumption on the number of 1-gons and 2-gons is here to ensure connectedness with high probability. Goal of the rest of the talk: present some results supporting this conjecture. Thomas Budzinski Random maps with unconstrained genus
Algebraic methods The conjecture is true if we only look at the sequence of degrees instead of the graph: for triangulations [Gamburd 2006] , for any ( P n ) as long as there is no 1-gon or 2-gon [Chmutov–Pittel 2013] . This implies # V ( M P n ) = ( 1 + o ( 1 )) log n , and convergence of the largest degrees to a Poisson–Dirichlet process. In both papers, the proof relies on algebraic methods (representations of the symetric group). Does not allow to say anything more precise about the graph. The assumption that there is no 1-gon or 2-gon does not look optimal. Thomas Budzinski Random maps with unconstrained genus
Probabilistic methods? Theorem (B.–Curien–Petri 2019) We assume that P n has o ( √ n ) one-gons and o ( n ) two-gons. Let v n 1 , v n 2 , . . . be the vertices of M P n , ranked by decreasing degrees. Let [ i , j ] n be the number of edges of M P n between v n i and v n j . Then � 1 � ( d ) 2 n deg ( v n i ) − n → + ∞ ( X i ) i ≥ 1 . − − − → i ≥ 1 � 1 � ( d ) 2 n [ i , j ] n − n → + ∞ ( X i X j ) i , j ≥ 1 , − − − → i , j ≥ 1 where X is a Poisson–Dirichlet process. Probabilistic proofs! Thomas Budzinski Random maps with unconstrained genus
Peeling! Let ρ be the root vertex. Goal of the rest of the talk: explain 1 why 2 n deg ( ρ ) converges to a uniform variable on [ 0 , 1 ] . Peeling arguments! For us, a peeling exploration of M P n will be an increasing sequence of "partial gluings" ( S i ) 0 ≤ i ≤ n of the polygons of P n . S 0 is just the collection of polygons P n (as on the left) with a root, and S n = M P n . The non-glued edges of S i form the boundary of S i . Thomas Budzinski Random maps with unconstrained genus
Peeling! To move from S i to S i + 1 , we choose a boundary edge A ( S i ) according to a peeling algorithm , and glue it where it has to be glued in M P n . Important: at each step i , conditionally on S i , the gluing of the non-glued edges is uniform, so the peeled edge is glued uniformly to one of the 2 ( n − i ) − 1 boundary edges. At each step, there are "true" vertices (in black) and "false" vertices (in white). Thomas Budzinski Random maps with unconstrained genus
Peeling cases (Episode 1) Case 1: gluing two edges on two different components. Case 2: gluing two edges on the same hole of the same component Thomas Budzinski Random maps with unconstrained genus
Peeling cases (Episode 2) Case 3: gluing two neighbour edges on the same hole. Case 4: gluing two edges on two different holes of the same component. Case 3 creates vertices, and Case 4 creates genus. Thomas Budzinski Random maps with unconstrained genus
Peeling algorithm and closure time Peeling algorithm: always peel the edge on the right of the root vertex. At the time τ where ρ becomes a true vertex, stop the exploration. The closure of the root occurs when the peeled edge is glued 1 to the edge on its left, so P ( τ = i + 1 | S 0 , . . . , S i ) = 2 ( n − i ) − 1 . Consequence: convergence in distribution 1 n τ → T , where T 1 has density 2 √ 1 − t d t . Thomas Budzinski Random maps with unconstrained genus
First and second moment estimates Let us fix a "blue" vertex v on a polygon at the beginning, and let θ be the time at which it is glued to the peeled vertex. For the same reason as τ , we have 1 1 n θ → Θ ∼ 2 √ 1 − t d t , with Θ independent of T . Hence P ( v glued to ρ | τ ) = 1 − √ 1 − τ . By summing over all vertices √ 1 ( P ) 2 n E [ deg ( ρ ) | τ ] − n → + ∞ 1 − − − − → 1 − τ. Same thing with two blue vertices gives � 1 � ( P ) � Var 2 n deg ( ρ ) � τ − n → + ∞ 0 . − − − → 1 Finally 2 n deg ( ρ ) converges in distribution to 1 − √ 1 − τ ∼ Unif ([ 0 , 1 ]) . Thomas Budzinski Random maps with unconstrained genus
Uncheating Peeling cases 5 and 6: We need show that peeling case 5 does not occur before time 1 − o ( 1 ) . For this, we needed to understand how 1-gons appear, and why they disappear quickly when we peel them. Thomas Budzinski Random maps with unconstrained genus
Geometric motivations By the uniformization theorem, this provides a way to build random hyperbolic Riemannian surfaces (i.e. constant curvature − 1) [Brooks–Makover 2004] . Peeling techniques allow to control the diameter of these surfaces [B.–Curien–Petri 2019+] . The surface we obtain has genus g ∼ n 2 and diameter ( 2 + o ( 1 )) log g . Question: are there hyperbolic surfaces with genus g and diameter ( 1 + o ( 1 )) log g ? Thomas Budzinski Random maps with unconstrained genus
THANK YOU ! Thomas Budzinski Random maps with unconstrained genus
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